MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
CHAPTER 3 Electrochemistry and Electrical Circuits and Their Elements
Read This Chapter to Learn About ➤ Electrostatics ➤ Electric Circuits ➤ Magnetism ➤ Electrochemistry ➤ Specialized Cells––Nerve Cells
ELECTROSTATICS
A simple static shock, the beating of the heart, the operation of household appliances, and the devastating damage inflicted by a lightning bolt—all of these examples in nature involve applications of electrostatics. Electrostatics is the study of electrically charged particles—their properties such as mass and charge, their behavior such as conservation of charge, and their interactions such as the repulsive or attractive forces that occur and the calculation of the magnitude of such forces through Coulomb’s law. This chapter reviews the fundamental concepts of electrostatics. Electric Charge and Charge Conservation Electric charge q is a physical property of the basic building blocks of the atom,a fundamental property of all matter. The SI unit of charge is the coulomb, abbrevi- ated C. Although charge can be positive or negative, the magnitude of charge is
53 MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
54 − UNIT I: e = 1.6 × 10 19 C. Considering the particles of the atom, the charge of the positively − Physical charged proton is +1.6 × 10 19 C, and the charge of the negatively charged electron is Foundations of − −1.6 × 10 19 C. Two like charges (either two positive charges or two negative charges) Biological Systems repel each other. Positive and negative charges attract each other. Electric charge is a conserved quantity and thus follows conservation of charge:
Electric charge can neither be created nor destroyed, only transferred. The net charge of a system remains constant.
Conductors and Insulators As you will read in the following section on electric circuits, it is important to identify materials that will either allow or prevent the flow of electric charge. Conductors such as metals are materials that allow the storage of or facilitate the flow of electric charge. Insulators such as rubber or wood prevent the storage or flow of electric charge.
Electric Force: Coulomb’s Law
Coulomb’s law describes the electrostatic force Fel between two charged particles q1 and q2, separated by a distance r:
q q q q = 1 1 2 = 1 2 Fel k 4πεo r2 r2
−12 2 2 where εo is the permittivity constant, defined as εo = 8.85 × 10 C /N · m . Values for k are:
1 N · m2 k = = 9.0 × 109 4πεo C2
Electrostatic force, as is the case for all types of forces, is a vector quantity and is expressed in units of newtons (N). The direction of the electrostatic force is based on the charges involved. Unlike charges generate an attractive (negative) force, and the direction is toward the other charge; like charges generate a repulsive (positive) force, and the direction is away from the other charge.
EXAMPLE: Determine the electrostatic force between two alpha particles of − − charge +2e (3.2 × 10 19 C) separated by 10 13 m.
SOLUTION: The electrostatic force can be determined using Coulomb’s law, · 2 . × −19 . × −19 = × 9 N m 3 2 10 C 3 2 10 C Fel 9 10 C2 1 × 10−13 m 2
− = 9.22 × 10 2 N, repulsive MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
55 CHAPTER 3: Electric Field Electrochemistry Electric field E defines the electric force exerted on a positive test charge positioned and Electrical Circuits and Their at any given point in space. A positive test charge, q , is similar in most respects to o Elements a true charge except that it does not exert an electrostatic force on any adjacent or nearby charges. Thus the electric field of a positive test charge provides an idealized distribution of electrostatic force generated by the test charge and is given by:
Fel E = qo Because E is a vector quantity, the direction is dependent on the identity of the charge. Because the test charge is positive, if the other charge is negative, an attractive force is generated and the direction of E is toward the negative charge. Likewise, if the other charge is positive, a repulsive force is generated and the direction of E is away from the positive charge. Electric field E is expressed in units of newtons per coulomb. If E is known, it is possible to determine the electrostatic force exerted on any charge q placed at the same position as the test charge using: = Fel qoE An electric field can be produced by one or more electric charges. The electric field of a point charge, which always points away from a positive charge and toward a nega- tive charge, can be calculated by direct substitution of Coulomb’s law into the expres- sion for E: qq k o Fel 2 q E = = r = k qo qo r2
ELECTRIC FIELD LINES
Electric field lines represent a visual display of the electric field that uses imaginary lines to represent the magnitude and direction of the electric field or the distribution of the electrostatic force over a region in space. The lines of force from a positive charge are directed away from the positive charge, whereas the lines of force of a negative charge are directed toward the negative charge, as depicted in Figure 3-1. The magni- tude of the force is greater in the region closer to the charge and becomes weaker as the distance from the charge increases.
ELECTRIC FIELD DUE TO CHARGE DISTRIBUTION
For more than one charge in a defined region of space, the total electric field Etot, because it is a vector quantity, is the vector sum of the electric field generated by each
charge Eq in the distribution, or = + + + +··· Etot Eq1 Eq2 Eq3 Eq4 MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
56
UNIT I: Repulsive Electrostatic Force of Like Charges Physical Foundations of Biological Systems +
+
Attractive Electrostatic Force of Unlike Charges
-
+
FIGURE 3-1 Electric field lines. Source: From George Hademenos, Schaum’s Outline of Physics for Pre-Med, Biology, and Allied Health Students, McGraw-Hill, 1998; reproduced with permission of The McGraw-Hill Companies.
Electric Potential The concept of potential in this regard is similar to the potential that was discussed in Chapter 1. It discussed the fact that potential energy becomes stored by an object as a result of work done to raise the object against a gravitational field. The electric poten- tial V at some point B becomes stored as a result of work done, W, against an electric field to move a positive test charge from infinity (point A) to that point (point B), or
−W V = qo work Electric potential = charge The electric potential is a scalar quantity that can be positive, negative, or zero, depending on the sign and magnitude of the point charge as well as the work done. Electric potential is expressed in units of joules per coulomb (J/C) = volts (V). The electric potential difference, V, between any two points A and B in an elec- tric field is related to the work done by the electrostatic force to move the charge from point A to point B as: −W = − = AB V VB VA qo MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
57 CHAPTER 3: Electrochemistry and Electrical Circuits and Their Elements
FIGURE 3-2 Equipotential surfaces generated for a positive charge.
The absolute electric potential at a point A that exists at a distance r from a charged particle at point B depends on the magnitude of charge at point B as well as the distance according to the following formula: q V = k r This is only true if one assumes that V → 0 at a point infinitely far away. Because the electric potential is a scalar quantity, the electric potential for n charges can be deter- mined by adding the electric potential values calculated for each of the n charges: n = = + + + +···+ V Vi Vq1 Vq2 Vq3 Vq4 Vqn i=1 Equipotential surfaces are a graphical method of representing the electric poten- tial of any charge distribution as concentric circles that are normal or perpendicular to electric field lines. Consider the example of a positive charge in Figure 3-2. The electric field vectors are pointed away radially in all directions from the charge. Equipoten- tial surfaces can also be drawn on the diagram to represent the electric potential of a positive charge at any distance r from the charge. Recall that the electric potential of a q point charge is given by V = k . On the surface of the charge where r is at its minimum, r the electric potential is at its greatest, and thus a solid circle is drawn about the point charge to represent the largest magnitude of the potential at the surface of the point charge. As the distance r increases, the electric potential is represented as concentric circles that become larger in circumference.
EXAMPLE: The Bohr model of the hydrogen atom describes electron motion in a − circular orbit of radius 0.53 Å (angstrom, where 1 angstrom = 1 × 10 10 m) about the nuclear proton. Determine the following for the orbiting electron:
1. The electric field 2. The electric potential MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
58 UNIT I: SOLUTION: Physical Foundations of 1. The electric field E experienced by the orbiting electron is: Biological Systems q E = k r2 − where q =−1.6 × 10 19 C and r is the distance between the electron and the proton. Substituting values yields: − N · m2 −1.6 × 10 19 C N E = 9 × 109 =−5.14 × 1011 C2 0.53 × 10−10 m 2 C
where the minus sign indicates that the electric field is directed toward the electron. 2. The electric potential V of the electron is: q V = k r Substituting values yields: − N · m2 1.6 × 10 19 C N · m V = 9 × 109 = 27.2 = 27.2V C2 0.53 × 10−10 m C
ELECTRIC DIPOLE
An electric dipole consists of two point charges, +q and −q, typically equal in magni- tude but opposite in sign separated by a small distance d, as shown in Figure 3-3.
FIGURE 3-3 Electric dipole.
The electric potential, V, can be found for the electric dipole through the superpo- sition of the two potentials: +q −q 1 1 r− − r+ V = V+q + V−q = k + k = kq − = kq r+ r− r+ r− r+r− MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
59 The objective of this discussion is to calculate the potential of the dipole at a point CHAPTER 3: much farther from the dipole than the distance of separation between the two charges. Electrochemistry With such conditions, the following approximations can be made: and Electrical Circuits and Their 1. r−r+ ≈ r2 Elements 2. r− − r+ ≈ d cos θ
Substituting these approximations into the equation for electric potential given previ- ously yields an expression for the potential of an electric dipole:
qd cos θ V = k r2 where the product qd is referred to as the physical quantity, dipole moment p. The dipole moment is a vector quantity, pointing in a direction from the negative charge to the positive charge, with units of coulomb meters (Cm). The equipotential surfaces of an electric dipole are shown in Figure 3-4.
FIGURE 3-4 Equipotential surfaces of an electric dipole.
ELECTRIC CIRCUITS
You do not have to look far to realize the significance of electric circuits: they are involved in providing light to a room and power to an appliance. There are two types of electric circuits: direct current (DC) circuits and alternating current (AC) circuits. This section will focus on DC circuits. In a DC circuit, electricity in the form of an electric charge generated by a voltage source (such as a battery) flows as current through an arrangement of circuit elements or devices (e.g., a lightbulb, an alarm, a motor) that are all connected by a conductor, or a material that allows electric charge to flow easily through it. These circuit elements may be connected in series or parallel, as shown in Figure 3-5. In a series circuit, two MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
60 UNIT I: or more circuit elements are connected in sequence, one after the other. In a parallel Physical circuit, two or more circuit elements are connected in a branching arrangement, such Foundations of that a charge may pass through one element or the other. Regardless of the type of cir- Biological Systems cuit, these elements interact with and can directly influence the flow of charge through the circuit.
FIGURE 3-5 Series circuit and parallel circuit.
CURRENT
Current, I, is the rate of motion of electric charge and is expressed as: electric charge moving through a region Current = time required to move the charge q I = t where current is measured in SI units of amperes, A.
➤ Current in a series circuit: Current that flows through one circuit element con- nected in series must also flow through the remaining elements connected in the series circuit. Therefore, the net or total current that flows through a series circuit remains the same through each of the individual elements or = = = =···= Inet I1 I2 I3 In
where n refers to the nth element in a series circuit. ➤ Current in a parallel circuit: In a parallel circuit, one circuit element branches into two or more connected elements. The current that flows from the voltage source MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
61 separates, and in the process divides its current among the branch elements. The CHAPTER 3: amount of current that enters each of the branch elements depends on the resis- Electrochemistry tance of that element. In any event, the net or total current in a parallel circuit is and Electrical Circuits and Their equal to the sum of the currents that flow through all connected branching ele- Elements ments, or
= + + +···+ Inet I1 I2 I3 In
The flow of current in a series and parallel circuit is illustrated in Figure 3-6.
FIGURE 3-6 Flow of current in a series and parallel circuit.
Voltage Electromotive force, E (often abbreviated emf), is the voltage or potential difference generated by a battery or power source when no current is flowing. When current is flowing through a circuit with conductive elements of resistance R, the battery gener-
ates an internal resistance Rint, resulting in a voltage drop through the battery of IRint. The voltage generated by an emf through the circuit is given by:
E − − = IRint IR 0
Electromotive force is expressed in units of volts (V).
➤ Voltage in series. The net voltage of a circuit with conductive elements connected in series is the sum of the individual voltages across each conductive element of the circuit or = + + +···+ Vnet V1 V2 V3 Vn
➤ Voltage in parallel. The net voltage of a circuit with conductive elements con- nected in parallel is identical to the individual voltages across each conductive element of the circuit or = = = =···= Vnet V1 V2 V3 Vn MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
62 UNIT I: Physical Resistance Foundations of Resistance, R, is the inherent property of a conductor by which it resists the flow of Biological Systems electric current and represents a measure of the potential difference V that must be supplied to a circuit to drive current, I, through the circuit. Resistance is defined as: potential difference Resistance = current V R = I Resistance is expressed in units of volts per ampere, also known as the ohm ().
OHM’S LAW
Ohm’s law states that the voltage V across a resistor R is proportional to the current I through it, and can be written in equation form as:
V = IR
An increase in voltage drives more electrons through the wire (conductor) at a greater rate; thus the current increases. If the voltage remains the same and the wire is a poor conductor, the resistance against the flow of electrons increases, thereby decreasing the current. It is possible to change I by manipulating V and R, but it is not possible to change V by manipulating I because current flow is due to the differ- ence in voltage, not vice versa. The current is a completely dependent variable.
EXAMPLE: The filament of a lightbulb has a resistance of 250 . Determine the current through the filament if 120 V is applied to the lamp.
SOLUTION: The current is related to the voltage by Ohm’s law. V 120 V I = = = 0.48 A R 250
As current flows through a resistor, electric power P is dissipated into the resistor according to the following equivalent expressions: V 2 P = IV = I 2 R = R The unit for electric power is the watt, abbreviated W.
RESISTORS IN SERIES
In a circuit consisting of three resistors connected in series, as in Figure 3-7, the cur- rent must flow through the path presented by the resistors in series. To simplify circuit
calculations, the equivalent resistance R eq can be calculated in terms of the resistance of the individual components: = + + +···+ R eq R1 R2 R3 Rn MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
63 CHAPTER 3: R R R 1 2 3 Electrochemistry and Electrical Circuits and Their R R R R eq = 1 + 2 + 3 Elements
FIGURE 3-7 Resistors in series. Source: From George Hademenos, Schaum’s Outline of Physics for Pre-Med, Biology, and Allied Health Students, McGraw-Hill, 1998; reproduced with permission of The McGraw-Hill Companies.
= = EXAMPLE: Two resistances R1 8 and R2 6 are connected in series. Determine the equivalent resistance.
SOLUTION: = + = + = R eq R1 R2 8 6 14
RESISTORS IN PARALLEL
In a circuit consisting of three resistors connected in parallel, as in Figure 3-8, the cur- rent must flow through the path presented by the resistors in parallel. To simplify cir-
cuit calculations, the equivalent resistance Req can be calculated in terms of the resis- tance of the individual components:
1 = 1 + 1 + 1 +···+ 1 R eq R1 R2 R3 Rn
where the equivalent resistance R eq is always less than the smallest value of resistance of the individual components.
R 1
R 2
R 3
1 1 1 1 R =++R R R eq 1 2 3 R R R R 1 2 3 eq = R1R2 + R1R3 + R2R3
FIGURE 3-8 Resistors in parallel. Source: From George Hademenos, Schaum’s Outline of Physics for Pre-Med, Biology, and Allied Health Students, McGraw-Hill, 1998; reproduced with permis- sion of The McGraw-Hill Companies.
EXAMPLE: A circuit with four resistances connected in parallel yields an equiva- = = = lent resistance of 1 .IfR1 5 , R2 5 , and R3 10 , determine R4. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
64 UNIT I: SOLUTION: Four resistances connected in parallel are related to the equivalent Physical resistance by: Foundations of 1 1 1 1 1 Biological Systems = + + + R eq R1 R2 R3 R 4
1 = 1 + 1 + 1 + 1 1 5 5 10 R 4 = Solving for R 4 yields R 4 2 .
RESISTIVITY
The resistance R of a conductor depends on the resistivity, ρ, unique to the material, its length, L, and its cross-sectional area, A,or L R = ρ A Resistivity is given in units of ohm · meter ( · m).
− EXAMPLE: Given that the resistivity of copper is 1.7 × 10 8 · m, determine the resistance of a copper wire of 0.30 millimeter (mm) in diameter and 5 m in length.
SOLUTION: The resistance of the copper wire is related to its diameter and length by:
L L − 5.0m R = ρ = ρ = (1.7 × 10 8 · m) = 1.2 π 2 − 2 A r 3.14 0.15 × 10 3 m
Capacitance Capacitors are circuit elements that store charge and consist typically of two con- ductors of arbitrary shape carrying equal and opposite charges separated by an in- sulator. Capacitance, C, depends on the shape and position of the capacitors and is defined as: q C = V where q is the magnitude of charge on either of the two conductors and V is the mag- nitude of potential difference between the two conductors. The SI unit of capacitance is the coulomb/volt, collectively known as the farad (F).
PARALLEL PLATE CAPACITOR
The most common type of capacitor is the parallel-plate capacitor consisting of two large conducting plates of area A and separated by a distance d. The capacitance of a parallel-plate capacitor is: A C = κεo d MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
65 −12 2 2 where κ is a dielectric constant (dimensionless) and εo = 8.85 × 10 C /N · m = CHAPTER 3: − 8.85 × 10 12 F/m. For vacuum, κ = 1. Electrochemistry and Electrical Circuits and Their ENERGY OF A CHARGED CAPACITOR Elements
Because capacitors store positive and negative charge, work is done in separating the two types of charge, which is stored as electric potential energy W in the capacitor, given by: 1 1 1 q2 W = qV = CV2 = 2 2 2 C where V is the potential difference and q is charge.
EXAMPLE: To restore cardiac function to a heart attack victim, a cardiac defibril- lator is applied to the chest in an attempt to stimulate electrical activity of the heart and restore the heartbeat. A cardiac defibrillator consists of a capacitor charged to approximately 7.5 × 103 V with stored energy of 500 watt seconds (W· s). Determine the charge on the capacitor in the cardiac defibrillator.
SOLUTION: The energy stored in a capacitor is 1 W = CV2 2 and the charge on the capacitor is
q = CV
Solving for C gives q C = V Substituting into the expression for W, you get 1 q 1 W = V 2 = qV 2 V 2 Solving for q results in 2W 2 · 500 W · s q = = = 0.13 C V 7.5 × 103 V
CAPACITORS IN SERIES
The effective capacitance Ceff of capacitors connected in series, as shown in Figure 3-9, is given by:
1 = 1 + 1 + 1 +···+ 1 Ceff C1 C2 C3 Cn
where Cn is the nth capacitor connected in series. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
66 UNIT I: C C C Physical 1 2 3 Foundations of Biological Systems 1 1 1 1 = ++ C C C C eff 1 2 3 C C C C 1 2 3 eff = C1C2 +C1C3 + C2C3
FIGURE 3-9 Capacitors in series. Source: From George Hademenos, Schaum’s Outline of Physics for Pre-Med, Biology, and Allied Health Students, McGraw-Hill, 1998; reproduced with permis- sion of The McGraw-Hill Companies.
C 1
C 2
C 3
C C C C eff = 1 + 2 + 3
FIGURE 3-10 Capacitors in parallel. Source: From George Hademenos, Schaum’s Outline of Physics for Pre-Med, Biology, and Allied Health Students, McGraw-Hill, 1998; reproduced with permis- sion of The McGraw-Hill Companies.
CAPACITORS IN PARALLEL
The effective capacitance Ceff of capacitors connected in parallel, as shown in Figure 3-10, is given by:
= + + +···+ Ceff C1 C2 C3 Cn
where Cn is the nth capacitor connected in parallel.
TABLE 3-1
Summary of Circuit Element Quantities of DC Circuits
Circuit Element Series Parallel = + + +···+ = = = =···= Voltage Vnet V1 V2 V3 Vn Vnet V1 V2 V3 Vn = = = =···= = + + +···+ Current Inet I1 I2 I3 In Inet I1 I2 I3 In = + + +···+ 1 = 1 + 1 + 1 +···+ 1 Resistance R eq R1 R2 R3 Rn R eq R1 R2 R3 Rn 1 = 1 + 1 + 1 +···+ 1 = + + +···+ Capacitance Ceff C1 C2 C3 Cn Ceff C1 C2 C3 Cn MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
67 DIELECTRICS CHAPTER 3: Electrochemistry A dielectric is an insulator that is usually used to fill the gap between the plates of a and Electrical capacitor because it increases the capacitance. Dielectric materials are characterized Circuits and Their Elements by a dielectric constant, κ, which relates the new capacitance of the capacitor to its original capacitance by the relation:
C = κCo
where Co is the capacitance of a capacitor with an empty gap and C is the capacitance of a capacitor filled with a dielectric. The dielectric constant is equal to 1 for vacuum and slightly greater than 1 for air (κ = 1.00054).
Conductivity Conductivity refers to the ability of an object or substance to conduct electricity. Conductivity is a property that can be applied to both metals (metallic conductivity) and substances (electrolytic solutions). Metallic conductivity is directly related to the movement of valence or free electrons that can easily be removed from the element’s outer shell. Under normal conditions, these valence electrons are randomly spinning in their position. When the element is connected within an electric circuit and sub- jected to an electrical power source, electric current exerts a push or force on the free electrons, causing them to move between the positive and negative terminals of the battery. In addition to electrons, ions or charged atoms within a solution become free to move under an electric current supplied by a battery. In electrolytic conductivity, the electric current causes the positively charged and negatively charged ions to move in opposite directions.
Conductivity Meters A conductivity meter is an instrument equipped with a probe designed to measure the conductivity of a substance within a solution. In performing measurements, as the probe is placed within the solution, an electric voltage is applied between two elec- trodes within the probe, with the electrical resistance that is provided by the solution causing a reduction in voltage. The voltage change is a reflection of the solution’s ionic strength and hence the conductivity of the substance.
MAGNETISM
The working principles of a television monitor, the physical properties of Earth and the basis for the most effective navigational tool in history (the compass) all have a common thread—magnetism. A property of matter discovered by the ancient Chinese MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
68 UNIT I: in the mineral lodestone, magnetism is involved in a wide spectrum of applications in Physical science and technology and represents an important subject in physics. This chapter Foundations of covers the basics of magnetism. Biological Systems
Magnetic Field Magnetic field B is an attractive or repulsive force field generated by a moving charged particle. For a standard bar magnet with a north pole (where the lines of force begin) and a south pole (where the lines of force end), the magnetic field is attractive for opposite poles and repulsive for like poles. Magnetic field is a vector quantity. The unit of magnetic field is the tesla (T), where newton weber 1 tesla = 1 = 1 · 2 ampere meter meter
Magnetic fields can also be measured in units of gauss (G), a unit used for measur- ing smaller magnetic fields. These two units are related through the conversion ratio 1T= 1 × 104 G. Magnetic fields are created in the presence of moving charges or current. You can calculate the magnitude of the magnetic field and determine its direction based on the behavior or configuration of the current. Current can flow through a long, straight wire; through a wire loop or coil; through a long solenoid (a long, straight cylinder consisting of many loops of wire wrapped around the cylinder); or through a toroid (a long, cylin- der of wire coils that is bent in the form of a doughnut). If current is flowing through each of these four configurations, the magnitude of the magnetic field can be calcu- lated according to the formulas noted in Figure 3-11, where the constant in each of the −7 equations, μo = 4π × 10 T · m/A, is termed the permeability of free space. The di- rection of the magnetic field can be determined using the right-hand rule, as depicted in the figure.
EXAMPLE: Earth can produce magnetic fields as high as 600 milligauss (mG). Express this value of magnetic field in terms of teslas.
SOLUTION: The units of tesla and gauss are related according to
1T= 104 G
Therefore, − 1T − 600 mG = 600 × 10 3 G = 6 × 10 5 T 104 G
EXAMPLE: A circular conducting coil of diameter 0.4 m has 50 loops of wire and a current of 3 A flowing through it. Determine the magnetic field generated by the coil. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
69 CHAPTER 3: Electrochemistry and Electrical Circuits and Their Elements
FIGURE 3-11 Calculating the magnitude of a magnetic field. Source: From Frederick J. Bueche and Eugene Hecht, Schaum’s Outline of College Physics, 10th ed., McGraw-Hill, 2006; reproduced with permission of The McGraw-Hill Companies.
SOLUTION: The equation for the magnetic field in the center of a circular loop or coil is − T · m 4π × 10 7 (50)(3A) μ NI A − B = o = = 4.7 × 10 4 T 2r 2 0.2m
Magnetic Force on a Charged Particle in Motion The magnitude of the magnetic force, F, or the force exerted on a charged particle q moving with a velocity v in a uniform magnetic field B is defined as:
F = qvB sin θ
where θ is the angle between the lines of the magnetic field B and the direction of the velocity v of the charged particle. The direction of the force can be determined by implementation of the right-hand rule. Given a charged particle moving with a velocity MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
70 UNIT I: Physical Foundations of Biological Systems
FIGURE 3-12 Using the right-hand rule to determine the direction of a force. Source: From George Hademenos, Schaum’s Outline of Physics for Pre-Med, Biology, and Allied Health Students, McGraw-Hill, 1998; reproduced with permission of The McGraw-Hill Companies.
q v perpendicular to a uniform magnetic field B, the right hand is positioned such that the thumb points in the direction of q v and the remaining four fingers are aligned in the direction of B. The direction of the magnetic force F is perpendicular to the palm, as illustrated in Figure 3-12.
Magnetic Force on a Current-Carrying Wire The magnitude of the magnetic force exerted on a current-carrying wire of current I of length L placed in a uniform magnetic field B is defined as
F = ILBsin θ
where L is the length of the wire (conductor) and θ is the angle between the current and the magnetic field. The direction of the magnetic force on the wire can be found by orienting the thumb of the right hand along the axis of the wire with the remaining fingers in the direction of the magnetic field. The magnetic force is directed upward from the aligned palm, as shown in Figure 3-13.
F I Current L Force S N
B
Magnetic field
FIGURE 3-13 Determining the magnetic force on a current-carrying wire. Source: From Arthur Beiser, Schaum’s Outline of Applied Physics, 4th ed., McGraw-Hill, 2004; reproduced with per- mission of The McGraw-Hill Companies.
EXAMPLE: A wire of length 40 centimeters (cm) carrying a current of 30 A is ◦ − positioned at an angle of 50 to a uniform magnetic field of 10.0 × 10 4 watts per square meter (W/m2). Determine the magnitude and direction of the force exerted on this wire. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
71 SOLUTION: The magnetic force F exerted on the wire of length L with current I CHAPTER 3: placed in a magnetic field B is given by: Electrochemistry and Electrical F = ILBsin θ Circuits and Their − − ◦ − = (30 A)(40 × 10 2 m)(10.0 × 10 4 W/m2)(sin 50 ) = 9.2 × 10 3 N Elements
Using the right-hand rule where the fingers are aligned with the magnetic field lines and the thumb is aligned with the current direction, the force is directed perpendicularly into the page.
ELECTROCHEMISTRY
Electrochemistry is the study of oxidation–reduction reactions. These are reactions in which electrons are transferred from one species to another. The species that gains electrons is reduced; the species that loses electrons is oxidized. A reduction– oxidation, or redox, reaction that is spontaneous produces electricity. A nonsponta- neous redox reaction requires electricity to run. Batteries (voltaic or galvanic cells) operate by producing electricity via a spontaneous redox reaction. Electrolytic cells require electricity to make the reaction occur; they are used in certain industrial pro- cesses, such as the purification of aluminum from its ore. These processes are called electrolysis reactions.
Electrolytic Cells ELECTROLYSIS
Electrolysis is the process of splitting or breaking up compounds to stimulate chemical change by passing electricity through the solutions involved in the process. Electrolytic cells are constructed of nonspontaneous redox reactions that require electricity to make them run. See Figure 3-14.
e– e–
Cathode Anode
FIGURE 3-14 Electrolytic cell. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
72 UNIT I: In an electrolytic cell, the cathode is negative and the anode is positive. Reduction Physical occurs at the anode, described by the half reaction Foundations of + + − → Biological Systems Y e Y and oxidation occurs at the cathode, described by the half-reaction − − Z → Z + e
FARADAY’S LAWS OF ELECTROLYSIS
Faraday’s laws of electrolysis quantify the electrolysis process by establishing a math- ematical relationship between the amount of elements deposited at an electrode or the gas liberated as a result of electric current that passes through the solution. The two laws are written as follows: Faraday’s First Law of Electrolysis. During electrolysis within a solution, the mass of any substance deposited or liberated at an electrode is directly proportional to the electric charge that passes through the solution. Faraday’s Second Law of Electrolysis. During electrolysis within a solution, the masses of different substances deposited or liberated at an electrode by the same amount of electric charge are proportional to the equivalent masses of the substances.
To calculate the current (in amperes) required to deposit a certain mass of metal in an electroplating experiment, you must have the following:
➤ Faraday constant, F ➤ n ➤ The time in seconds ➤ The molar mass of the metal
EXAMPLE: Calculate the current needed to deposit 365 milligrams (mg) silver in 216 minutes (min) from aqueous silver ion.
SOLUTION: + − ➤ From Ag 1 → Ag is a 1e change. ➤ Start with the gram amount, and convert it by steps into amperes. ➤ A = C/s. − 1 mole Ag 1 mole e 9.65 × 104 C (0.365g Ag) − 107.9 g Ag 1 mole Ag mole e 1 × = 0.0251 A 1.30 × 104 s
EXAMPLE: Calculate the mass of iodine formed when 8.52 milliamperes (mA) − flows through a cell containing I for 10.0 min. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
73 SOLUTION: CHAPTER 3: − − ➤ From I to I is a total of 2e change. Electrochemistry 2 and Electrical ➤ Start with the amp and convert it by steps into g I2. Circuits and Their ➤ A × s = C. Elements − − 1 mole e 1 mole I2 254 g I2 (8.52 × 10 3 A)(600 s) 4 − 9.65 × 10 C 2 mole e mole I2 = . × −3 6 73 10 gI2
Concentration Cell A concentration cell is a type of electrolytic cell that is composed of two half-cells with the same electrodes separated by a salt bridge but different concentrations. Upon acti- vation, the cell acts to dilute the more concentrated solution and strengthen the more dilute solution. During the process, electrons are transferred from the lower concen- tration cell to the higher concentration cell, creating a voltage as it approaches equi- librium. The potential of the concentration cell can be determined quantitatively by the Nernst equation:
◦ RT E = E − ln Q nF where R = gas constant = 8.315 J/K mol F = Faraday constant [products]coefficient Q = reaction quotient = [reactants]coefficient ◦ E = energy produced by reaction T = temperature in Kelvin n = number of electrons exchanged in balanced redox equation E = cell potential
Voltaic Cells Voltaic cells (also called galvanic cells) are constructed so that a redox reaction pro- duces an electric current. A diagram of a voltaic cell is shown in Figure 3-15. One half-cell is the anode. This is where the oxidation takes place. The electrode wears away as the reaction proceeds, with the metal electrode becoming an ion in the solution. The other half-cell is the cathode. This is where the reduction is taking place. The electrode builds up as the reaction proceeds, with the metal ions in the solution plating out as pure metal. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
74 UNIT I: Physical Foundations of Biological Systems
Cathode
Anode
+ + FIGURE 3-15 Voltaic cell with Zn/Zn 2 anode and Cu/Cu 2 cathode.
The two half-cells are connected by a salt bridge between the solutions and by a wire between the electrodes. The anode gives up electrons: + − Zn (s) → Zn 2 (aq) + 2e
The cathode uses the electrons: + − Cu 2 (aq) + 2e → Cu (s)
BALANCING OXIDATION–REDUCTION REACTIONS
Balancing a redox equation is not the same as balancing a nonredox equation. You have to take the electron change into account.
How to Balance a Redox Equation in an Acidic Solution
1. Divide the equation into two half-reactions, one for oxidation and one for reduction. 2. Balance all the atoms.
➤ Start with the main atoms. ➤ Then use H2O to balance O. + ➤ Then use H to balance H.
3. Determine the total electron change for each half-reaction, using the oxidation numbers of each species that is changing. 4. Determine the lowest common denominator (LCD) of the total electron change. 5. Multiply through each half-reaction so that the electron change equals the LCD. 6. Add the two half-reactions. 7. Cancel where applicable. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
75 How to Balance a Redox Equation in a Basic Solution CHAPTER 3: Electrochemistry 1. Balance as for an acidic solution. and Electrical + 2. Change all H to H2O. Circuits and Their − 3. Put the same number of OH on the other side. Elements 4. Add and cancel where applicable.
EXAMPLE IN ACID SOLUTION:
−2 + −2 → +3 + Cr2 O7 C2O4 Cr CO2 (The phase labels have been left out for clarity.)
SOLUTION: ➤ Divide into two half-reactions. −2 → +3 Cr2O7 Cr −2 → C2O4 CO2
➤ Balance the atoms, starting with the main atoms. −2 → +3 Cr2O7 2Cr −2 → C2O4 2CO2 ➤ Use H2O to balance the oxygen atoms. −2 → +3 + Cr2O7 2Cr 7H2O −2 → C2O4 2CO2 + ➤ Use H to balance the hydrogen atoms. + + −2 → +3 + 14H Cr2O7 2Cr 7H2O −2 → C2O4 2CO2
➤ Determine the oxidation number of each atom that is changing and find the total electron change for each half-reaction. ➤ Cr on the left is +6; Cr on the right is +3. This is a change of 3 electrons, but there are 2 Crs changing, so the total electron change is 6 for this half- reaction. ➤ C on the left is +3; C on the right is +4. This is a change of 1 electron, but there are 2 Cs changing, so the total electron change is 2 for this half- reaction. ➤ Find the LCD of the electron change. The LCD of 6 and 2 is 6. ➤ Balance the electrons; the electron change for both half-reactions must be 6. ➤ The first half-reaction is already a change of 6 electrons, so it stays as is. The second half-reaction is a change of 2, so it must be multiplied by 3.
+ + −2 → +3 + − = 14H Cr2O7 2Cr 7H2O e 6 −2 → − = 3C2O4 6CO2 e 6 MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
76 UNIT I: ➤ Sum the two half-reactions. Physical + + −2 + −2 → +3 + + Foundations of 14H Cr2O7 3C2O4 2Cr 7H2O 6CO2 Biological Systems ➤ There is no canceling to do, so this is the final balanced equation.
EXAMPLE IN BASIC SOLUTION:
+2 + → + Co H2O2 Co(OH)3 H2O (The phase labels have been left out for clarity.)
SOLUTION:
➤ Divide into two half-reactions.
+2 → Co Co(OH)3 → H2O2 H2O ➤ + Balance the atoms, using H2O to balance oxygen, and H to balance hydrogen.
+ +2 → + + 3H2O Co Co(OH)3 3H + + → 2H H2O2 2H2O
➤ Determine the electron change for each species that is changing.
➤ Co is +2 on the left; it is +3 on the right; this is a 1 electron change, and there is 1 Co atom changing, so the total electron change is 1. ➤ Ois−1 on the left, it is −2 on the right; this is a 1 electron change, but there are 2 oxygen atoms changing, so the total electron change is 2.
➤ The LCD of 1 and 2 is 2. Therefore the n value is 2.
➤ The first half-reaction must be multiplied by 2.
+ +2 → + + 6H2O 2Co 2Co(OH)3 6H + + → 2H H2O2 2H2O
➤ Sum the two half-reactions.
+ +2 + + + → + + + 6H2O 2Co 2H H2O2 2Co(OH)3 6H 2H2O
➤ Cancel out where the same species shows up on both sides.
+ +2 + → + + 4H2O 2Co H2O2 2Co(OH)3 4H ➤ + − Change all H to H2O’s and put the same number of OH on the other side. − + + +2 + → + 4OH 4H2O 2Co H2O2 2Co(OH)3 4H2O
➤ Cancel again.
− + +2 + → 4OH 2Co H2O2 2Co(OH)3 MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
77 CELL NOTATION FOR VOLTAIC CELLS CHAPTER 3: Electrochemistry Rather than draw the voltaic cell, a notation can be used instead. Cell notation is drawn and Electrical as follows: Circuits and Their Elements anode | anode’s ion ||cathode’s ion | cathode
The single line indicates a phase change; the double line indicates the salt bridge between the half-cells. The salt bridge is necessary to complete the electrical circuit and allow the reaction to take place. The concentration of the ions is often written directly after the ion. The cell notation for the cell in Figure 3-12 looks like: + + Zn | Zn 2 ||Cu 2 | Cu
It is read as follows: At the anode + Zn (s) → Zn 2 (aq)
At the cathode + Cu 2 (aq) → Cu (s)
The sum + + Zn (s) + Cu 2 (aq) → Cu (s) + Zn 2 (aq) Sometimes a redox reaction produces a gas or another species that is an ion in solution. In this case, the anode or cathode, being in solution, or being a gas, cannot have a wire attached. A platinum electrode is used to complete the circuit.
EXAMPLE: + + + Zn (s) + 2Fe 3 (aq) → Zn 2 (aq) + 2Fe 2 (aq) + + + Zn | Zn 2 ||Fe 3,Fe 2 | Pt
STANDARD CELL POTENTIAL ◦ ◦ Ecell is the symbol given to the cell potential under the standard conditions of 25 C, 1 M ◦ concentrations for all solutions, and 1 atm pressure for all gases. The Ecell is calculated from the table of reduction potentials, as shown on next page.
TABLE OF REDUCTION POTENTIALS
The table of reduction potentials is used as follows:
➤ This is a table of reductions. ➤ For the oxidation half-reaction (it is above the reduction half-reaction in the table), the reaction is reversed and the sign of the value is reversed. ➤ For a spontaneous reaction, the anode reaction is above the cathode reaction. ➤ The anode reaction is the reverse of the reduction reaction. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
78
UNIT I: TABLE 3-2 Physical ◦ Foundations of Standard Electrode Reduction Potentials in Aqueous Solution at 25 C Biological Systems Reduction Half-Reaction Standard Potential (V)
Li+ (aq) + e− → Li (s) −3.04 Ba+ (aq) + e− → Ba (s) −2.71 Mg+2 (aq) + 2e− → Mg (s) −2.38 Al+3 (aq) + 3e− → Al (s) −1.66 Zn+2 (aq) + 2e− → Zn (s) −0.76 Cr+3 (aq) + 3e− → Cr (s) −0.74 Fe+2 (aq) + 2e− → Fe (s) −0.41 Cd+2 (aq) + 2e− → Cd (s) −0.40 Ni+2 (aq) + 2e− → Ni (s) −0.23 Sn+2 (aq) + 2e− → Sn (s) −0.14 Pb+2 (aq) + 2e− → Pb (s) −0.13 Fe+3 (aq) + 3e− → Fe (s) −0.04 + + − → ( ) 2H (aq) 2e H2 g 0.00 Sn+4 (aq) + 2e− → Sn+2 (aq) 0.15 Cu+2 (aq) + e− → Cu+ (aq) 0.16 Cu+2 (aq) + 2e− → Cu (s) 0.34 + − → − I2 (s) 2e 2I (aq) 0.54 Fe+3 (aq) + e− → Fe+2 (aq) 0.77 Ag+ (aq) + e− → Ag (s) 0.80 ( ) + − → − Br2 l 2e 2Br (aq) 1.07 ( ) + + + − → ( ) O2 g 4H (aq) 4e 2H2O l 1.23 −2 + + + − → +3 + ( ) Cr2O7 (aq) 14H (aq) 6e 2Cr (aq) 7H2O l 1.33 ( ) + − → − Cl2 g 2e 2Cl (aq) 1.36 − + + + − → +2 + ( ) MnO4 (aq) 8H (aq) 5e Mn (aq) 4H2O l 1.49 + + + − → ( ) H2O2 (aq) 2H (aq) 2e 2H2O l 1.78 −2 + − → −2 S2O8 (aq) 2e 2SO4 (aq) 2.01 ( ) + − → − F2 g 2e 2F (aq) 2.87
➤ Never change the values except for the sign. ➤ ◦ Add the two values to give Ecell. ➤ The weakest reduction reactions are at the top of the table. (They have the most negative potential.) ➤ The strongest reduction reactions are at the bottom of the table. (They have the most positive potentials.) ➤ ◦ The Ecell must always be positive for a spontaneous reaction that produces a volt- age.
◦ EXAMPLE: Determine Ecell for the following cell: | | +3 || | − | Al Al I2 I Pt MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
79 SOLUTION: CHAPTER 3: Electrochemistry ➤ The anode is an oxidation. The table of reduction potentials gives the value and Electrical + −1.66 V for the reduction half-reaction of Al 3 to Al. Therefore, the value for Circuits and Their the oxidation half-reaction is +1.66 V. Elements ➤ The cathode is the reduction. The table of reduction potentials gives the value +0.54 V for this half-reaction. ➤ + . ◦ Adding the two values gives 2 2 V for the Ecell. It is positive, as it must be for a spontaneous reaction.
STANDARD CELL POTENTIAL AND STANDARD FREE ENERGY CHANGE ◦ ◦ The G can be calculated from Ecell using the following equation: ◦ =− ◦ G nFEcell ◦ ◦ For a spontaneous reaction, Ecell is positive, and G is negative. ◦ EXAMPLE: Calculate the G for the cell given previously.
➤ +3 − − − = Al to Al is a 3e change; I2 to I is a 2e change; the n 6. ◦ ➤ Plug in the values and calculate G :
◦ G =−(6)(9.65 × 104 J/V)(2.2V) =−1.3 × 106 J
EQUILIBRIUM CONSTANT AND STANDARD CELL POTENTIAL ◦ The equilibrium constant for a reaction can be calculated from Ecell using the equation ◦ = ( . / ) Ecell 0 0257 n ln K = ◦ = . EXAMPLE: For the previous cell, with n 6 and Ecell 2 19 V, the equilibrium constant is:
2.2 = (0.0257/6) ln K
513 = ln K
K = e513 = too large to calculate
NONSTANDARD CELL POTENTIALS
The Nernst equation is used to calculate the nonstandard cell potential; this occurs when the concentrations are other than 1 M. The Nernst equation is given as:
= ◦ − ( . / ) Ecell Ecell 0 0257 n ln Q MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
80 UNIT I: The equation requires the following: Physical ➤ Foundations of A balanced equation Biological Systems ➤ The Q expression and the value of Q ➤ The n value ➤ ◦ The value of Ecell ➤ ◦ The values of Ecell, n, and Q to be plugged in to calculate the nonstandard Ecell
EXAMPLE: Calculate the nonstandard cell potential for the cell + + Al | Al 3 (0.150 M) ||Zn 2 (0.075 M) | Zn − − 3e 2e
SOLUTION:
➤ Determine the n value which in this case is 6. ➤ Write the balanced equation. + + 2Al (s) + 3Zn 2 (aq) → 2Al 3 (aq) + 3Zn (s)
➤ Write the Q expression and calculate Q. +3 2 ( . )2 = [Al ] = 0 150 = . Q + 53 6 [Zn 2]3 (0.075)3 ➤ ◦ Calculate the Ecell. + Al → Al 3 +1.66 + Zn 2 → Zn −0.76 +0.90
➤ Plug into the Nernst equation. = . − ( . / ) . Ecell 0 90 0 0257 6 ln 53 6 = . Ecell 0 88 V
Batteries ELECTROMOTIVE FORCE OR VOLTAGE
The electromotive force, emf, of a cell, or the cell voltage generated by the reaction,
is called Ecell. If you know the Ecell (by measuring it in the lab), you can calculate the work, w, produced by the cell using the equation w =− max nFEcell − − where n = LCD of the mole e , F = 9.65 × 104 coulombs (C)/mole e , and 1 joule (J) = 1 volt-coulomb (V-C). MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
81 = . EXAMPLE: Calculate the work done by the cell as shown next if the Ecell 0 650 V CHAPTER 3: Electrochemistry | +2 | || | +| Pt Hg2 Hg (l) H2 H Pt and Electrical Circuits and Their SOLUTION: Elements ➤ Calculate the n value.
= +2 → − n 2 because Hg2 Hg is a 2e change → + − H2 2H is a 2e change ➤ w = (− ) . × 4 =− . × 5 max 2 (9 65 10 C) (0.650 V) 1 25 10 J ➤ The sign is negative because work is produced.
Examples of voltaic cells include: (1) lead-storage batteries for cars, which use Pb
for an anode, PbO2 for a cathode, and H2SO4 as an electrolyte, and (2) nickel-cadmium batteries, which are drycell batteries where the anode and cathode are rechargeable.
SPECIALIZED CELLS—NERVE CELLS
The functional unit of the nervous system is the neuron (see Figure 3-16). The neu- ron is a highly specialized cell, which contains all of the organelles typically found in eukaryotic cells. It is highly suited for communication because of its wirelike projec- tions, known as dendrites, which carry impulses toward the central cell body. The cell body is a thicker region of the neuron containing the nucleus and most of the cyto- plasm. The axon is a projection, generally very long, that carries impulses away from the cell body. A typical neuron has a single axon, which may combine with other axons to form a single nerve. The neurons are supported by glial cells, which are often referred to as neuroglia if located in the brain and spinal cord. In the outlying neurons of the peripheral nervous system, which carries impulses to and from the central nervous system, the support- ing tissue consists of Schwann cells. Schwann cells tend to grow around the axon so that it is wrapped in a multilayered insulating cover called the myelin sheath. This fatty, membranous sheath allows for a rapid and highly efficient “insulated wire” for impulse transmission. The myelin sheath is regularly interrupted along the length of the neuron by short stretches of unmyelinated membrane, referred to as the nodes of Ranvier. When a neuron is not conducting an impulse, it is said to be in its resting state.In this condition, a resting potential or a difference in charge exists between the inside and outside of the membrane. A higher concentration of sodium ions exists outside the membrane, while there is a higher concentration of potassium ions inside. In addition, a number of negatively charged proteins reside on the inside. These concentration MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
82 UNIT I: Dendrite Physical Foundations of Biological Systems
Cell body
Node of Ranvier Axon
Myelin sheath
Synaptic knobs
FIGURE 3-16 A model of a typical motor neuron. Source: From Sylvia S. Mader, Biology, 8th ed., McGraw-Hill, 2004; reproduced with permission of The McGraw-Hill Companies.
gradients are maintained by two factors: the impermeability of the resting membrane + + + + to Na and the action of a Na /K pump that, driven by ATP, transfers Na to the + outside and pumps K inside. Because of these gradients, the inside of the neuron is negative relative to the outside; a potential difference of approximately −60 millivolts exists across the membrane. The natural tendency to correct this energetically unsta- ble imbalance is the driving force behind the nerve impulse. When a neuron is stimulated, the point of stimulation suddenly becomes perme- able to sodium ions, which rush in, depolarizing the membrane as the incoming posi- + tive ions balance the negative internal charge. Enough Na rushes in to actually make the inside of the membrane positive for a few milliseconds. This shift of charge constitutes the neural impulse,oraction potential. Although it occurs at only one place on the neuron, it triggers a depolarization of the adjacent area, thus initiating a new action potential. This process continues as a wave of depo- larization down the length of the axon. The impulse is thus not actually transported anywhere but like a wave of water is re-created at each point. At any point on the neuron when the action potential reaches a maximum (about +40 mV) of the interior relative to the exterior, the membrane suddenly again becomes + + impermeable to Na . At the same time, K is pumped out, until it essentially balances MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2
83 the number of sodium ions that rushed in and the membrane is repolarized. This CHAPTER 3: efflux of positive ions restores the resting potential of −60 millivolts (albeit with potas- Electrochemistry + + sium ions rather than sodium ions). After the resting potential is established, the Na /K and Electrical Circuits and Their pumps restore the original sodium and potassium gradients existing before the initia- Elements tion of the action potential. Until the membrane reaches its resting potential again, it is incapable of developing a new action potential; while this is the case, the membrane is said to be in its refractory period. MCAT-3200184 book November 13, 2015 14:38 MHID: 1-25-958837-8 ISBN: 1-25-958837-2